funorbfandomcom-20200223-history
Talk:Dungeon Assault
Formulas Could someone check the formula for the stone golem because according to my calculations the probability of beating it is ~33% with a raider with 10 in everything making it the strongest room not factoring in special abilities. Ok forget that I had the attack and defence the wrong way round Also wouldn't the formula for the follow rooms be different: *Crossive slime - Because of the Sneak vs. Detect roll & the fighting *Demon lord - Due to fleeing Also wouldn't the special abilities on some raiders effect their formulas? Chao.master 13:47, 3 November 2008 (UTC) :Yes, it would be different. I don't think we need to write a new formula for the Demon Lord, and the Corrosive Slime maybe, maybe not. We do, however, still need a new formula for the Black Knight. I was at one point working on coding a program for automatically finding out the probability based on the raider, room, and any special abilities or stat effects. However, I sorta forgot about it and haven't worked on it for a few weeks... maybe I'll get back to it. In the meantime, if you want, you could make the other formulas. -TimerootTalk • • 23:29, 3 November 2008 (UTC) ::I would be quite happily calculate the formulas but I'm not sure how, I would ask my teacher but I got work xp next week. I knew how you arrived at the current formulas I could work them out :::I used the formula for an infinite geometric series, with some assumptions about how the probabilities are calculated from the raider's and room's stats. If you want me to go into more detail I will. Quartic ~ insanity is a virtue | Talk 20:45, 8 November 2008 (UTC) :::: That would be nich thx. Chao.master 12:09, 9 November 2008 (UTC) Derivation of the formula for Advancing Let: * 0\leq a\leq 1 be the probability of winning an attack roll. (raider attack vs. room defence) * 0\leq d\leq 1 be the probability of winning a defence roll. ( 0\leq d_f\leq 1 in the main article) * 0\leq s\leq 1 be the probability of winning a sneak roll. then the probability of winning after n \geq 0 rounds of combat, where the raider won the sneak roll is: p_n = (1-a)^nd^na The probability that the raider wins after n \geq 0 rounds of combat, where the raider lost the sneak roll is: \tilde{p}_n = (1-a)^nd^{n+1}a This gives the probability of winning as: p = s\sum^\infty_{n=0} p_n + (1-s)\sum^\infty_{n=0} \tilde{p}_n = s\frac{a}{(1-d(1-a))} + (1-s)\frac{ad}{(1-d(1-a))} = \frac{as + ad(1-s)}{(1-d(1-a))} The other equations can be derived in a similar manner. Quartic ~ insanity is a virtue | Talk 16:38, 10 November 2008 (UTC) Program? Maybe we should make a program that calculates all these success-thing things? ^^ :Are you suggesting a program that players download and run on their computer? Or are you suggesting having something embedded in the page? I can see potential problems with both. Quartic ~ insanity is a virtue | Talk 19:10, 15 November 2008 (UTC) ::Oh sorry for not signing. I was thinking about an embed version ~ [[User:Ad Fundum|'Ad Fundum']]Talk ~ 19:11, 15 November 2008 (UTC) :::I thought that was most likely. The problem I can see with it is whether it is actually possible. I've noticed that the RuneScape Wiki has created some calculators, but these require the user to edit and preview the page, which isn't ideal. Quartic ~ insanity is a virtue | Talk 19:21, 15 November 2008 (UTC) ::::It was just a rough idea that popped into my head :P If it is not possible, then nvm ^^ ~ [[User:Ad Fundum|'Ad Fundum']]Talk ~ 19:23, 15 November 2008 (UTC) Weakness of Chaos Champion What is the weakness of a chaos champion? a little other question: Does corrisive slime has to be defeaten by offensive or disarming? ~ [[User:Ad Fundum|'Ad Fundum']]Talk ~ 15:10, 17 November 2008 (UTC) :A Chaos Champion is more susceptible to monsters than traps due to traps requiring two failed rolls to actually defeat the raider. It is more likely to die from a monster than a Black Knight is, due to the Black Knight's ability, and thus is best for disarming traps. It is great, but it's not perfect. Remember that waking the dragon is enough for it to deem you unworthy, and you have to pay the full price to get it back! 19118219 Talk 18:30, 18 November 2008 (UTC) ::But more monsters in my dungeon make it very easy for people with 2 black knights.. ~ [[User:Ad Fundum|'Ad Fundum']]Talk ~ 20:17, 18 November 2008 (UTC) Dungeon Assault Calculator I decided to get back to work on the dungeon assault calculator which had lain stagnant for several weeks.... It's almost done, all I need is aproval from JAGeX. However, that's taking them a while, despite my bumping.... anyway, I saw that you can't upload .exe files to the Wiki. So, I thought of two possible solutions: *Convert to JavaScript, and somehow embed it. I don't know how to do that on a wiki like this, but I think it can be done, and it would look nice. *Read the bits to a .txt file, post it online, that way, anyone could just copy and paste the text in a new .txt file, rename it as a .exe file, and run. Not sure if this would work, but it's a very simple program, without any dependencies, so it should. Tell me what you think. And of course, this is all assuming that JAGeX allows it... -TimerootTalk • • 06:42, 7 December 2008 (UTC) :Hehe, that's what i suggested a few post above this :D I would love to see it! But why does Jagex have to aprove it? ~ [[User:Ad Fundum|'Ad Fundum']]Talk ~ 08:34, 7 December 2008 (UTC) ::I think the Javascript option is better and the only other option would be convert it to Java but I don't think we can embed java applets here anyway. [[User:D_P60|'DP ~ *Insert witty remark here*']] 10:26, 7 December 2008 (UTC) :::JS would have to be protected - only modifiable by admins - to prevent people doing things which lock up a browser or spawn millions of popups. The other option would be to use an external site (signed applet, Javascript, or server-side scripting) which read its rules from the Wiki. OrbFu 12:58, 7 December 2008 (UTC) ::::I got back to thinking about this... We do already have a downloadable version, but an embedded type would be nice. It should be possible to put something in Common.js that would make this work, correct? The problem with that is the fact that the MediaWiki software doesn't allow tags. (See? I didn't need to type for that to work!) So, the other option would be a template, taking the raider as a first parameter, the room as a second, and returning a little box, with a picture of each, and the probabilities listed. If anyone can think of a better way to take input, please post here. Timeroot Talk • • 18:09, 10 March 2009 (UTC) :::::I can see two routes to a proper interface: either write an extension (see the extension for the basics) or use JavaScript's ability to modify the DOM to insert the elements you want. The latter is probably more straightforward but less robust. OrbFu 20:33, 10 March 2009 (UTC) ::::::Proof of concept is working in User:OrbFu/Sandbox. Can other people see User:OrbFu/global.js or do I need to copy the contents somewhere? Also I'd note that it's probably best, if possible, to have a separate MediaWiki:da_calc.js and include that rather than put the entire calculator in Common.js. In fact, ideally Common.js would load da_calc.js conditionally upon the page title so that we don't slow down page loads for everything else in the wiki. OrbFu 23:04, 10 March 2009 (UTC) :::::::Okay, I finally got it working. That is some extremely cool JavaScript - man, I'll never understand the DOM. Meameamealokkapoowa oompa kudos to you. Anyway, I'll try to start on making a rough version of the program to demonstrate. And has it occured to anyone yet that this the potential to also make a great damage calculator for aog? Timeroot Talk • • 0 Attack What were to happen if a raider, with an attack stat of 0, went up against a monster, also with an attack stat of 0. This would be possible if the raider had been killed by rot worms, specters, etc. numerous times, and hadn't been healed. The room could be a monster with 0 attack if it was, say, giant rats, and there was a zealot in the raider's party. What would happen? Since none of them could kill each other, would they just go into eternal combat? Personally, I think zealots are pretty lousy raiders, so I've never encountered this. Does anyone know, though? TimerootTalk • • 01:08, 9 December 2008 (UTC) :I believe that if one participant has a stat of 0 then they automatically lose. I'm not sure what happens if a stat of 0 faces off against a stat of 0. OrbFu 23:38, 10 December 2008 (UTC) ::Yeah when someone reaches 0 in a stat they die. However I'm pretty sure that attack is the last thing to be calculated so it would be their sneak or whatever verse the opponents sneak thing. But it would still continue as normal because you can have 1 atk and so can the opponent and it's random who wins. You can have 2 atk and the opponent has 1 and the opponent can still win. [[User:D_P60|'DP ~ *Insert witty remark here*']] 01:44, 11 December 2008 (UTC) :::4/4 times my spy with zero attack beat rot worms/giant rats with zero attack (I had a zealot.) The screen was just filled up with the usual stuff (raider attacks, monster attacks), and at the end it said "raider wins!" I made a video of this also. If you want me to do it again I can but my zero attack spy goes bye bye on january 7th. PeaceBear0 03:38, 1 January 2009 (UTC) Archives The archives for this page have a non-standard naming system - the lack the blank inbetween "Archive" and the number. So that the archive template works here, could an admin move them for me? TimerootTalk • • 00:45, 28 December 2008 (UTC) :Done. Quartic ~ insanity is a virtue | Talk 01:04, 28 December 2008 (UTC) Why is my name there? In the list of the titles, it has my name before them. I want to know why my name is there, or if there is just some program that automatically puts your name in somehow... I have never edited this page, so I really want to know who did.-- Long Live Armadyl 22:25, 9 July 2009 (UTC) Ok, I got on my other computer and found that it has "" where my name was. I find that very odd.-- Long Live Armadyl 22:27, 9 July 2009 (UTC)